Problem 33 - Basic Technical Mathematics with Calculus

Question

Use the factor theorem and synthetic division to determine whether or not the second expression is a factor of the first. $$2 x^{5}-x^{3}+3 x^{2}-4 ; \quad x+1$$

Step-by-Step Solution

Answer

No, \( x+1 \) is not a factor of \( 2x^5 - x^3 + 3x^2 - 4 \).

Step 1: Understanding the Factor Theorem

The factor theorem states that a polynomial \( f(x) \) has a factor \( x - k \) if and only if \( f(k) = 0 \). In this problem, \( x+1 \) can be rewritten as \( x - (-1) \). Thus, we need to evaluate the polynomial at \( x = -1 \).

Step 2: Substituting \( x = -1 \) Into the Polynomial

Substitute \( x = -1 \) into the polynomial \( 2x^5 - x^3 + 3x^2 - 4 \). Calculate each term separately: \( 2(-1)^5 = -2 \), \( -(-1)^3 = 1 \), and \( 3(-1)^2 = 3 \). Sum these with the constant term: \(-2 + 1 + 3 - 4 = -2\).

Step 3: Analyzing the Result

Since \( f(-1) = -2 \) is not equal to zero, according to the factor theorem, \( x+1 \) is not a factor of the polynomial \( 2x^5 - x^3 + 3x^2 - 4 \).

Step 4: Confirming with Synthetic Division

Perform synthetic division with \( x = -1 \). Write the coefficients: [2, 0, -1, 3, 0, -4]. Carry down the first coefficient (2), then multiply by -1 and add consecutively: [2, \(-1\times2 + 0 = -2\), \(-1\times(-2) -1 = 1\), \(-1\times1 + 3 = 2\), \(-1\times2 + 0 = -2\), \(-1\times(-2) -4 = -2\)]. The remainder is -2, confirming \( x+1 \) is not a factor.

Key Concepts

Synthetic Division

Synthetic division is a simplified way to perform polynomial division, specifically when dividing a polynomial by a linear factor. It's a favored method due to its simplicity and efficiency. When you use synthetic division, you deal solely with the coefficients of the polynomial, making the process less tedious than regular polynomial long division. To begin, you organize the coefficients in descending order of the polynomial's degree. Missing terms are represented with a zero.

• Arrange the coefficients of the polynomial in descending order.
• Use the root of the divisor (e.g., solve \(x + 1 = 0\), which gives \(x = -1\)) to perform the division.
• Bring down the leading coefficient.
• Multiply the root by the number just written down and add the next coefficient repeatedly through each step.

By the end, the synthetic division provides you the quotient's coefficients and the remainder. If the remainder is zero, the divisor is a factor. In our example, the remainder is \(-2\), confirming \(x+1\) is not a factor.

Polynomial Division

Polynomial division, in principle, is similar to long division with numbers. This technique involves dividing a polynomial, called the dividend, by another polynomial, the divisor, to yield a quotient polynomial and possibly a remainder. The aim is to simplify complex polynomial expressions or determine factors. If the remainder after division is zero, the divisor perfectly divides the polynomial, indicating it is a factor.

• The dividend is the polynomial you are dividing.
• The divisor is the polynomial by which you divide.
• The quotient is the result of the division.
• The remainder is what is left if the divisor doesn't divide evenly.

In our case, instead of using full polynomial division, synthetic division was used, which showed \(x+1\) didn't completely divide the polynomial, given a non-zero remainder. This underscores that \(x+1\) is not a factor of the original polynomial.

Polynomial Factors

Polynomial factors are expressions that multiply together to form a polynomial. Investigating factors is essential to understanding polynomial behavior and solving polynomial equations. The process simplifies polynomials into types or combinations. Usually, these factors are linear, like \(x - k\), where \(k\) is a root of the polynomial.
When checking if a polynomial has a specific factor, you can use the Factor Theorem. It states a polynomial \(f(x)\) has a factor \(x - k\) if \(f(k) = 0\). This theorem is crucial for checking potential factors without performing the division. If you find that substituting \(k\) into \(f(x)\) gives zero, \(x - k\) is a factor. In our exercise, since substituting \(-1\) for \(x\) resulted in \(-2\), \(x + 1\) wasn't a factor of \(2x^5 - x^3 + 3x^2 - 4\).

• Factors are crucial in breaking down polynomials into simpler forms.
• Correct factorization helps in solving polynomial equations.
• The Factor Theorem provides a quick method to check divisibility.

Understanding polynomial factors aids in graphing, integration, and further algebraic manipulation.